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Elanso [62]
3 years ago
13

To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to

the other. This is ;
(a) an observational study.
(b) a matched pairs experiment.
(c) a completely randomized experiment.
(d) a block design, but not a matched pairs experiment.
(e) impossible to classify unless more details of the study are provided
Engineering
2 answers:
Kruka [31]3 years ago
7 0

Answer:

(b) a matched pairs experiment.

Explanation:

Matched pairs experiment are a special case of a random block design and is usually used when the experiment only has two conditions. In this case the conditions are the different fertilizers.

Oduvanchick [21]3 years ago
3 0

Answer: (b) a matched pairs experiment.

Explanation:

A match pair design or experiment is a randomized experimental procedure. It can be used in the experiment with two different sets of conditions and the subjects can be grouped into pairs. It is based on some of the blocking variable. The subjects are randomly assigned to different treatments.  

The given situation is an example of a matched pair experiment. This is because of the fact that there are two sets of conditions that is the use of fertilizer A and fertilizer B on different halves of the plot randomly to determine their effects.

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Answer:

True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.

Option: A

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3 years ago
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1:c 2:False

Explanation:

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A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
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Answer:

10.8\ \text{lb/ft^2}

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Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

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From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

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Answer:

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